This is regarding article in this link which explains Kalman filter. In this article at one point in equation 4 author says
If we multiply every point in a distribution by a matrix A, then what happens to its covariance matrix Σ?
Well, it’s easy. I’ll just give you the identity:
How can derive this equation?
On
I have a small addition: The OP has not specified that $x$ is zero mean.
In general, ${\rm COV}[x]$ is defined as $\mathbb{E}[(x - \mathbb{E}[x])(x - \mathbb{E}[x])^T]$. If $x$ is zero mean, i.e., $\mathbb{E}[x]=0$, Omnomnomnom has already given the answer. If it is not, it still works, like this:
$$\begin{align}{\rm COV}[A x] & = \mathbb{E}[(Ax - \mathbb{E}[Ax])(Ax - \mathbb{E}[Ax])^T] \\ & = \mathbb{E}[(Ax - A\mathbb{E}[x])(Ax - A\mathbb{E}[x])^T ]\\ & = \mathbb{E}[A(x - \mathbb{E}[x])(x - \mathbb{E}[x])^T A^T ] \\ & = A \mathbb{E}[(x - \mathbb{E}[x])(x - \mathbb{E}[x])^T ]A^T \\ & = A {\rm COV}[x] A^T \\ \end{align}$$
We have $$ Cov(Ax) = \Bbb E((Ax)(Ax)^T) = \Bbb E(A(xx^T)A^T) = A\Bbb E(xx^T) A^T = A Cov(x)A^T $$